LOJ#137. 最小瓶颈路 加强版(Kruskal重构树 rmq求LCA)

题意

三倍经验哇咔咔

#137. 最小瓶颈路 加强版

#6021. 「from CommonAnts」寻找 LCR

#136. 最小瓶颈路

Sol

首先可以证明，两点之间边权最大值最小的路径一定是在最小生成树上

考虑到这题是边权的最大值，直接把重构树建出来

然后查LCA处的权值即可

输入文件过大，需要用RMQ算法求LCA

// luogu-judger-enable-o2
#include<bits/stdc++.h>
const int MAXN = 1e6 + 10;
using namespace std;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N, Q, S, tot, dfn[MAXN], rev[MAXN], dep[MAXN], id[MAXN][21], lg2[MAXN], rd[MAXN];
vector<int> v[MAXN];
void dfs(int x, int fa) {
    dfn[x] = ++tot; dep[x] = dep[fa] + 1; id[tot][0] = x; 
    for(int i = 0, to; i < v[x].size(); i++) {
        if((to = v[x][i]) == fa) continue;
        dfs(to, x);
        id[++tot][0] = x;
    }
}
void RMQ() {
	for(int i = 2; i <= tot; i++) lg2[i] = lg2[i >> 1] + 1;
    for(int j = 1; j <= 20; j++) {
        for(int i = 1; (i + (1 << j) - 1) <= tot; i++) {
            int r = i + (1 << (j - 1));
            id[i][j] = dep[id[i][j - 1]] < dep[id[r][j - 1]] ? id[i][j - 1] : id[r][j - 1];
        }
    }
}
int Query(int l, int r) {
    if(l > r) swap(l, r);
    int k = lg2[r - l + 1];
    return dep[id[l][k]] < dep[id[r - (1 << k) + 1][k]] ? id[l][k] : id[r - (1 << k) + 1][k];
}
int main() {
	freopen("a.in", "r", stdin);
    N = read(); Q = read(); S = read();
    for(int i = 1; i <= N - 1; i++) {
        int x = read(), y = read();
        v[x].push_back(y); v[y].push_back(x);
    }
    dfs(S, 0);
    RMQ();
    while(Q--) {
        int x = read(), y = read();
        printf("%d
", Query(dfn[x], dfn[y]));
    }
    return 0;
}